a, \(M=1+5+5^2+5^3+..+5^{29}\)

\(=\left(1+5\right)+5^2\left(1+5\right)+...+5^{28}\left(1+5\right)\)

\(=6+5^2.6+...+5^{28}.6=6\left(1+5^2+...+5^{28}\right)⋮6\)( đpcm )

a, 1+5+5^2+...+5^29

=(1+5)+(5^2+5^3)+...+5^28+5^29)

=(1+5)+5^2(1+5)+...+5^28(1+5)

=6+5^2*6+...+5^28*6

=6(5^2+...+5^28) chia hết cho 6

b, cậu xem lại đề hộ tớ nhaxem chia hết cho 32 hay 31

a) M = 1+ 5 +5^2+5^3+....+5^29 có 30 số chia thành 15 cặp mỗi cặp 2 số

= (1+ 5)+5^2(1+5)+.....+5^28(1+5)

= 6.(1 +5^2+...+5^28) chia hết cho 6

b) lỗi hả bạn

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Ta có \(\left(29^m+1\right)\left(29^m+2\right)\left(29^m+3\right)\left(29^m+4\right)\)

 \(\Rightarrow29^m\left(1+2+3+4\right)=29^m\cdot10⋮5\)

= 29 ^m +1 x 29^m+2 x 29^m+3 x 29^m+4

= 29^m x (1+2+3+4)

=29^mx10 chia hết cho 5

=> 29^m + 1 x 29^m + 2 x 29^m + 3 x 29^m + 4 chia hết cho 5

2.Gọi số cần tìm là \(x\left(x\ne0,x>9\right)\)

Ta có:

\(53=mx+2\left(m\in N\right)\\ \Rightarrow51=mx\\ \Rightarrow x\inƯ\left(51\right)\left(1\right)\\ 77=nx+9\left(n\in N\right)\\ \Rightarrow68=nx\\ \Rightarrow x\inƯ\left(68\right)\left(2\right)\)

Từ (1) và (2) ta có:

\(x\inƯC\left(51,68\right)\)

\(51=3\cdot17\\ 68=2^2\cdot17\\ \Rightarrow\text{ƯCLN}\left(51,68\right)=17\\ ƯC\left(51,68\right)=Ư\left(17\right)=\left\{1;17\right\}\)

Vì x > 9 nên x = 17

Vậy số chia là 17

3. Làm câu b trước, các câu kia trả lời tương tự hoặc áp dụng điều đã chứng minh

b,

\(a+a^2+a^3+a^4+...+a^{29}+a^{30}\\ =\left(a+a^2\right)+\left(a^3+a^4\right)+...+\left(a^{29}+a^{30}\right)\\ =a\left(1+a\right)+a^3\left(1+a\right)+...+a^{29}\left(1+a\right)\\ =\left(1+a\right)\left(a+a^3+...+a^{29}\right)⋮a+1\)

Vậy \(a+a^2+a^3+a^4+...+a^{29}+a^{30}⋮a+1\) với a thuộc N